Question

If $H_2+\frac{1}{2}{O}_2\to H_{2}O;\Delta H=-68.09kcal$
$K+H_{2}O+water\to KOH(aq.)+\frac{1}{2}{H}_2;\Delta H=-48.0kcal$
$KOH+water\to KOH\left(aq\right);\Delta H=-14.0kcal$
the heat of formation of $KOH$ is:

• A. $-68.39+48-14.0$
• B. $-68.39-48.0+14.0$
• C. $+68.39-48.0+14.0$
• D. $+68.39+48.0-14.0$

Answer 1

The correct option is B $-68.39-48.0+14.0$

(i) $H_2+\frac{1}{2}{O}_2\to H_{2}O;{\Delta H}_f=-68.09Kcal$

(ii) $K+H_{2}O+water\to KOH\left(aq\right)+\frac{1}{2}{H}_2;{\Delta H}_f=-48Kcal$

(iii) $KOH+water\to KOH\left(aq\right);{\Delta H}_f=-14.0Kcal$

(i) + (ii) - (iii) gives heat of formation of $KOH$

$\therefore$ Heat of formation of $KOH=(-68.09-48.0+14.0)Kcal$